# Equivalence Classes proof.. simple but no idea where to start...

• Nov 24th 2009, 02:45 PM
richmond91
Equivalence Classes proof.. simple but no idea where to start...
Prove that for elements x, y, z Zm, we have (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)

and that x ⊗ y = y ⊗ x

Zm being an equivance class [x]m for some x ∈ Z
• Nov 24th 2009, 03:01 PM
Plato
Quote:

Originally Posted by richmond91
Prove that for elements x, y, z Zm, we have (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) and that x ⊗ y = y ⊗ x
Zm being an equivance class [x]m for some x ∈ Z

I am so sorry to tell you this, but I have no idea what that question means.
The I cannot even read the notation.
Can you rewrite the question?
Or at least explain what the notation means.
• Nov 24th 2009, 03:18 PM
Sampras
Quote:

Originally Posted by richmond91
Prove that for elements x, y, z Zm, we have (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)

and that x ⊗ y = y ⊗ x

Zm being an equivance class [x]m for some x ∈ Z

Is $x \oplus y$ defined as $x+y-xy$?
• Nov 24th 2009, 03:23 PM
richmond91
sorry if it is not very clear... the question is regarding equivalence classes congruence classes of modulo m. does that make it any clearer? i'm not too sure what the question wants me to do..

its basically asking to prove that when adding and multiplying congruent classes in modular arithmetic it does not matter about where brackets are when adding, and the order of multiplication... i think.

Thanks

Ps - in Zm i mean Zm (set of integers modulo m)
• Nov 24th 2009, 03:38 PM
Sampras
Quote:

Originally Posted by richmond91
Prove that for elements x, y, z Zm, we have (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)

and that x ⊗ y = y ⊗ x

Zm being an equivance class [x]m for some x ∈ Z

So $[x]_{m} \oplus [y]_m = [x+y]_m$.